Boris and His Amazing Haircut

By xiaruize

思路

先排除存在 ai<bia_i<b_i 的情况，这种情况一定为 NO

考虑对于每一个在 bb 中出现的数，求出至少需要几把这个高度的剪刀

对于高度 hh ，设当前考虑 bx=by=hb_x=b_y=h

当 max⁡i=xybi≤h\max\limits^{y}_{i=x}b_i \leq h 时，显然 bxb_x 与 byb_y 可以共用一个， 否则需要分开使用两个

可以使用 ST表 计算区间最大值 当然你写线段树也没人拦着。。。

求出最小个数后检查是否足够即可

Code

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/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define debug(x) cerr << #x << ": " << x << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int t;
int n;
int a[N], b[N];
int m;
int c[N];
int f[N][22];
map<int, int> mp, cnt, sum;
// bool en;

void rmq_init()
{
    for (int i = 1; i <= n; i++)
        f[i][0] = b[i];
    int k = 20;
    for (int j = 1; j <= k; j++)
        for (int i = n; i >= 1; i--)
        {
            if (i + (1 << (j - 1)) <= n)
                f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
        }
}

int rmq(int i, int j)
{
    int k = log2(j - i + 1);
    return max(f[i][k], f[j - (1 << k) + 1][k]);
}

void solve()
{
    mp.clear();
    cnt.clear();
    sum.clear();
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
        cin >> b[i];
    rmq_init();
    cin >> m;
    for (int i = 1; i <= m; i++)
    {
        cin >> c[i];
        sum[c[i]]++;
    }
    for (int i = 1; i <= n; i++)
    {
        if (a[i] < b[i])
        {
            NO;
            return;
        }
    }
    for (int i = 1; i <= n; i++)
    {
        if (a[i] == b[i])
            continue;
        if (!mp[b[i]])
            mp[b[i]] = i;
        else
        {
            int l = mp[b[i]];//上一个同样的值
            int x = rmq(l, i);//区间最大值
            if (x > b[i])//不能共用
                cnt[b[i]]++;
            mp[b[i]] = i;
        }
    }
    for (auto v : mp)
    {
        if (v.sec)
            cnt[v.fir]++;
    }//最后一段没有统计
    for (auto v : cnt)
    {
        if (sum[v.fir] < v.sec)
        {
            NO;
            return;
        }
    }
    YES;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    // auto t_1=chrono::high_resolution_clock::now();
    cin >> t;
    while (t--)
        solve();
    // auto t_2=chrono::high_resolution_clock::now();
    // cout <<". Elapsed (ms): " << chrono::duration_cast<chrono::milliseconds>(t_2 - t_1).count() << endl;
    return 0;
}